<div dir="ltr"><div>Matt</div><div> </div><div>It is not return, but the bind the one that does the miracle of multiplication. </div><div>By its definition for the list monad, it applies the second term once for each element are in the first term. </div>
<div>So return is called many times. At the end, bind concat all the small lists generated</div></div><div class="gmail_extra"><br><br><div class="gmail_quote">2013/7/20 Matt Ford <span dir="ltr"><<a href="mailto:matt@dancingfrog.co.uk" target="_blank">matt@dancingfrog.co.uk</a>></span><br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">Hi All,<br>
<br>
I thought I'd have a go at destructing<br>
<br>
[1,2] >>= \n -> [3,4] >>= \m -> return (n,m)<br>
<br>
which results in [(1,3)(1,4),(2,3),(2,4)]<br>
<br>
I started by putting brackets in<br>
<br>
([1,2] >>= \n -> [3,4]) >>= \m -> return (n,m)<br>
<br>
This immediately fails when evaluated: I expect it's something to do<br>
with the n value now not being seen by the final return.<br>
<br>
It seems to me that the return function is doing something more than<br>
it's definition (return x = [x]).<br>
<br>
If ignore the error introduced by the brackets I have and continue to<br>
simplify I get.<br>
<br>
[3,4,3,4] >>= \m -> return (n,m)<br>
<br>
Now this obviously won't work as there is no 'n' value. So what's<br>
happening here? Return seems to be doing way more work than lifting the<br>
result to a list, how does Haskell know to do this? Why's it not in the<br>
function definition? Are lists somehow a special case?<br>
<br>
Any pointers appreciated.<br>
<br>
Cheers,<br>
<span class="HOEnZb"><font color="#888888"><br>
--<br>
Matt<br>
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<br></blockquote></div><br><br clear="all"><br>-- <br>Alberto.
</div>